Question

How do you find the derivative of `y=ln((x-1)/(x+1))^(1/3)`?

Answer 1

`dy/dx = 2/(3x^2-3)`

Explanation:

`y = ln((x-1)/(x-1))^(1/3)`
`:. y = 1/3ln((x-1)/(x+1))`
`:. 3y = ln((x-1)/(x+1))`
`:. e^(3y) = ((x-1)/(x+1))` ..... [1]

Differentiating the LHS implicity and the RHS using the quotient rule gives:

`:. 3e^(3y)dy/dx = ( (x+1)(1) - (x-1)(1) )/(x+1)^2`
`:. 3e^(3y)dy/dx = ( x+1-x+1 )/(x+1)^2`
`:. 3e^(3y)dy/dx = 2/(x+1)^2`

`:. 3((x-1)/(x+1))dy/dx = 2/(x+1)^2` (using [1])
`:. dy/dx = 2/3*1/(x+1)^2*(x+1)/(x-1)`
`:. dy/dx = 2/3*1/(x+1)*1/(x-1)`
`:. dy/dx = 2/3*1/(x^2-1)`

Hence,

`dy/dx = 2/(3x^2-3)`

Answer 2

`y = ln((x-1)/(x-1))^(1/3)`

`= 1/3ln((x-1)/(x-1))`

`= 1/3[ln(x-1)-ln(x+1)]`

Recall that `d/dx(lnu) = 1/u (du)/dx`, so

`d/dx(ln(x-1)) = 1/(x-1)` and `d/dx(ln(x+1)) = 1/(x+1)`

So

`dy/dx = 1/3[1/(x-1)-1/(x+1)]`

`= 2/(3(x^2-1))`